# Fourier transformation with the similarity theorem

I'm trying to calculate a simple fourier transformation but I got confused at a part and still don't know why .

$$f(t) = e^{−at^2}$$ knowing that $$F(w) = F({e^{−t^2}}) = \sqrt{\pi} . e^{-w^2/4}$$

We can calulate the Fourier transformation of f(t) . From the fourier table the result should be : $$\sqrt{\pi/a} . e^{-w^2/4a} (I)$$

But when calculating the function with the similarity theorem $$(1/|a|) . F(w/a)$$ I get : $$(\sqrt{\pi}/a) . e^{-w^2/4a^2}$$ which is not the same as ( I ) what I'm doing wrong here ?

isn't $$1/|a| F(w/a) = (1/a) . e^{-(w^2/a^2)/4} = 1/|a| . e^{-w^2/4a^2}$$ ?

It looks like you are using $$a^2$$ instead of $$a$$, i.e., you're computing $$F(e^{-(at)^2})$$ instead of $$F(e^{-at^2})$$.
It is not too hard to see what the result should be using a change of variable $$s = \sqrt{a}\,t$$ in the integral defining the FT (note your normalization of where the factors of $$\pi$$ or $$\sqrt{\pi}$$ go in the definition of the FT may be different; I was lazy, so I used this convention—I didn't bother to check if this definition of the FT works on Gaussians like it does in your post, but that doesn't matter, the important thing is that the change of variable approach works no matter what your normalizations in the definition of the FT are):
\begin{align*} F(e^{-at^2})(\omega) &= \int e^{-at^2}e^{i\omega t}\,dt\\ &= \int e^{-s^2}e^{i\omega(s/\sqrt{a})}\,d(\frac{s}{\sqrt{a}})\\ &= \frac{1}{\sqrt{a}}F(\omega/\sqrt{a})\\ &= \sqrt{\frac{\pi}{a}}\,e^{-(\omega/\sqrt{a})^2/4}\\ &= \sqrt{\frac{\pi}{a}}\,e^{-\omega^2/4a}. \end{align*}
If we plug in $$a^2$$ instead of $$a$$, we see we would get at the end $$F(e^{-a^2t^2})(\omega) =\sqrt{\frac{\pi}{a^2}}\,e^{-\omega^2/4a^2} = \frac{\sqrt\pi}{a}\,e^{-\omega^2/4a^2},$$ which is what you were getting.
• @Gaston: I updated my answer showing the relevant computation. As to the edit you made in your original post, I am not sure what you mean about $(1/|a|)F(w/a)$ being equal to $e^{-w^2/4a^2}$, you lost the factor of $1/|a|$ in the second equality. – Alex Ortiz Dec 8 '20 at 19:57